The Ratio Test

A geometric series is easy to judge: each term is the same multiple $r$ of the one before, and the series converges exactly when $|r| < 1$ . The ratio test asks whether a series behaves like a geometric series far out, and if so, with what ratio.

The idea: a ratio that settles

For $\sum \frac{n^2}{3^n}$ , the ratio of a term to the one before is

$\frac{a_{n+1}}{a_n} = \frac{(n+1)^2}{3^{n+1}} \cdot \frac{3^n}{n^2} = \frac{(n+1)^2}{3n^2} = \frac{1}{3}\left(1 + \frac{1}{n}\right)^2.$

At $n = 1$ the ratio is $\frac{4}{3}$ , so the second term is larger than the first. But as $n$ grows the ratio settles toward $\frac{1}{3}$ (the figure). Far out, each term is roughly a third of the one before, just like a geometric series with $r = \frac{1}{3}$ , and that series converges.

Dots at height (n + 1) squared over 3 n squared for n = 1 to 15, the ratio of consecutive terms of the series of n squared over 3 to the n. The first dot is at 4/3, above a solid line at 1; the rest fall below it and approach a dashed line at 1/3. The limit 1/3 is below 1, so the series converges. — **Figure 1.** The ratio of consecutive terms settles at $\frac{1}{3}$ , below the threshold 1. LevlPrep original diagram.

For a series $\sum a_n$ with nonzero terms, find

$L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|.$

Limit	Conclusion	Why
$L < 1$	the series converges	the terms shrink at least as fast as a convergent geometric series
$L > 1$ , or the limit is infinite	the series diverges	the terms grow, so they do not approach 0
$L = 1$	no conclusion	both convergent and divergent series give 1

The absolute value lets the test work on series whose terms change sign. Only the limit matters: any number of early ratios above or below 1 change nothing.

Simplifying the ratio

Most of the work is algebra. Write $a_{n+1}$ by replacing every $n$ with $n + 1$ , then multiply by the reciprocal of $a_n$ . Three simplifications come up again and again:

$\frac{(n+1)!}{n!} = n + 1$ and $\frac{(2n+2)!}{(2n)!}$ $= (2n+2)(2n+1)$ : two new factors, not one.
$\frac{5^{n+1}}{5^n} = 5$ .
$\frac{(n+1)^k}{n^k}$ $= \left(1 + \frac{1}{n}\right)^k \to 1$ : polynomial factors contribute 1 to the limit.

Worked example 1. Does $\sum_{n=1}^{\infty} \frac{5^n}{n!}$ converge?

$\frac{a_{n+1}}{a_n}$ $= \frac{5^{n+1}}{(n+1)!} \cdot \frac{n!}{5^n}$ $= \frac{5}{n+1}$ .

$L$ $= \lim_{n \to \infty} \frac{5}{n + 1}$ $= 0 < 1$ , so the series converges by the ratio test.

The first few ratios ( $\frac{5}{2}$ , $\frac{5}{3}$ , $\frac{5}{4}$ ) are above 1, and the terms grow at first. That does not matter; the limit does.

Worked example 2. Does $\sum_{n=1}^{\infty} \frac{n!}{2^n}$ converge?

$\frac{a_{n+1}}{a_n}$ $= \frac{(n+1)!}{2^{n+1}} \cdot \frac{2^n}{n!}$ $= \frac{n + 1}{2}$ , which grows without bound.

The limit is infinite, so the series diverges by the ratio test. (The terms themselves grow without bound, so the nth term test agrees.)

Worked example 3. What does the ratio test say about $\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}$ ?

$\frac{a_{n+1}}{a_n}$ $= \frac{n + 1}{(n+1)^2 + 1} \cdot \frac{n^2 + 1}{n}$ . The top and bottom both grow like $n^3$ , and the limit is 1.

No conclusion. Use limit comparison with $\frac{1}{n}$ instead: the limit of $\frac{n^2}{n^2 + 1}$ is 1, and the harmonic series diverges, so this series diverges.

When to reach for it

Use the ratio test when the terms contain factorials or exponentials such as $4^n$ , since those cancel cleanly in the ratio. For a $p$ -series or a ratio of polynomials the limit is always 1, so a comparison test is the right tool. Later in the unit the ratio test does its most important job: finding where a series with an $x$ in it converges.

Writing it the way graders want

Show the ratio with $a_{n+1}$ on top and the absolute value, simplify it, write the limit with $n \to \infty$ , and compare the value with 1 in words: " $L = 0 < 1$ , so the series converges by the ratio test." Do not say the series converges to $L$ : the limit of the ratios decides whether the series converges, not what its sum is.